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Excess of KI and dil. H(2)SO(4) were mix...

Excess of KI and dil. `H_(2)SO_(4)` were mixed in `50 mL H_(2)O_(2)`. The liberated `I_(2)` required 20 mL of `0.1 N Na_(2)S_(2)O_(3)`. Find out the strength of `H_(2)O_(2)` in g per litre.

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The correct Answer is:
0.68 g
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Excess of Kl and dill. H_(2)SO_(4) were mixed in 50 mL H_(2)O_(2) . Thus, l_(2) liberated requires 20 mL of 0.1 N Na_(2)S_(2)O_(3) . What will be the strenght of H_(2)O_(2) in g L^(-1) ?

25 mL of H_(2)O_(2) solution were added to excess of acidified solution of KI . The iodine so liberated required 20 mL of 0.1N Na_(2)S_(2)O_(3) for titration Calculate the strength of H_(2)O_(2) in terms of normalility, percentage and volumes. (b) To a 25 mL H_(2)O_(2) solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3N sodium thiosulphate solution. Calculate the volume strength of H_(2)O_(2) solution.

50 mL of an aqueous solution of H_(2)O_(2) was treated with an excess of KI solution and dilute H_(2)SO_(4) . The liberated iodine required 20 mL 0.1 N Na_(2)S_(2)O_(3) solution for complete interaction. Calculate the concentration of H_(2)O_(2) in g//L .

To a 25 mL of H_(2)O_(2) solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3 N Na_(2)S_(2)O_(3) solution. Calculate the volume strength of H_(2)O_2 solution. Strategy : Volume strength of H_(2)O_(2) solution is related to its normality by the following relation Volume strength (V)=5.6xx"Normality" (N) where, Normality =((meq)H_(2)O_(2))/V_(mL) According to the law of equivalence (meq)_(Na_(2)S_(2)O_(3))=(meq)_(I_(2))=(meq)_(H_(2)O_(2))

When 100 mL of an aqueous of H_(2)O_(2) is titrated with an excess of KI solution in dilute H_(2)O_(2) , the liberated I_(2) required 50 mL of 0.1 M Na_(2) S_(2)O_(3) solution for complete reaction. Calculate the percentage strength and volume strength of H_(2)O_(2) solution.

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